Problem: The sequences of positive integers $1,a_2, a_3, \dots$ and $1,b_2, b_3, \dots$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$. There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$. Find $c_k$.
Solution: Let $d$ be the common difference, and let $r$ be the common ratio, so $d$ and $r$ are positive integers.  Then $a_n = 1 + (n - 1) d$ and $b_n = r^{n - 1},$ so
\begin{align*}
1 + (k - 2) d + r^{k - 2} &= 100, \\
1 + kd + r^k &= 1000.
\end{align*}Then
\begin{align*}
(k - 2) d + r^{k - 2} &= 99, \\
kd + r^k &= 999.
\end{align*}From the second equation, $r^k < 999.$  If $k \ge 4,$ then $r < 999^{1/4},$ so $r \le 5.$

Since the geometric sequence is increasing, $r \neq 1,$ so the possible values of $r$ are 2, 3, 4, and 5.  We can write the equations above as
\begin{align*}
(k - 2) d &= 99 - r^{k - 2}, \\
kd &= 999 - r^k.
\end{align*}Thus, $99 - r^{k - 2}$ is divisible by $k - 2,$ and $999 - r^k$ is divisible by $k.$

If $r = 2,$ then the only possible values of $k$ are 4, 5, 6, 7, and 8.  We find that none of these values work.

If $r = 3,$ then the only possible values of $k$ are 4, 5, and 6.  We find that none of these values work.

If $r = 4,$ then the only possible values of $k$ is 4.  We find that this value does not work.

If $r = 4,$ then the only possible values of $k$ is 4.  We find that this value does not work.

Therefore, we must have $k = 3,$ so
\begin{align*}
d + r &= 99, \\
3d + r^3 &= 999.
\end{align*}From the first equation, $d = 99 - r.$  Substituting, we get
\[3(99 - r) + r^3 = 999,\]so $r^3 - 3r - 702 = 0.$  This factors as $(r - 9)(r^2 + 9r + 78) = 0,$ so $r = 9,$ so $d = 90.$  Then $a_3 = 1 + 2 \cdot 90 = 181$ and $c_3 = 9^2 = 81,$ and $c_3 = 181 + 81 = \boxed{262}.$